# Why does increasing subshell cause ionization energy to decrease?

I know that increasing subshell decreases ionization energy; for example, going from Beryllium to Boron. I do not understand why this is.

The answer my textbook (Chemistry 2e on OpenStax) gives is that the higher subshells (eg $$\ce{p}$$) has more energy than the lower subshells (eg $$\ce{s}$$). I do not understand why a higher energy subshell require more energy to remove (ie more ionization energy)? Is it as if the higher energy subshell is less stable, and thus has lower ionization energy?

• Yeah, that doesn't make sense. Ionization energy does increase in going from Be to B, but it's not because you're going to a higher-energy subshell (which you are), it's because you're staying in the same shell but increasing the effective nuclear charge. Could you provide the exact quote from your textbook? Aug 4, 2021 at 2:48
• 2p electrons have higher energy ( and lower ionization energy ) than 2s electrons within the same atom context, but these energies shift in the different element atom context. So 2s/2p( if excited ) electrons in Be have higher energies ( lower ionization energies ) than respective 2s/2p electrons in B. Aug 4, 2021 at 8:04